CUET Preparation Today
If $x=\frac{1}{t^2}$ and $y=\frac{1}{t^3}$, then $\frac{d^2y}{dx^2}$ at t = 1 is : |
$\frac{3}{2}$ $\frac{1}{2}$ $\frac{2}{3}$ $\frac{3}{4}$ |
$\frac{3}{4}$ |
$x=\frac{1}{t^2},y=\frac{1}{t^3}$ [Given] $\frac{dx}{dt}=-2\frac{1}{t^3}$ $\frac{dy}{dt}=-3\frac{1}{t^4}$ $=\frac{dy}{dx}=\frac{3}{2t}$ |
